Respuesta :
[tex]6|p|+8\ \textgreater \ -22[/tex]
Subtract 8. [tex]6|p|\ \textgreater \ -30[/tex]
Divide by 6. [tex]|p|\ \textgreater \ -5[/tex]
The solution set is all real numbers because no matter what number is chosen for p, its absolute value will be bigger than any negative number.
Subtract 8. [tex]6|p|\ \textgreater \ -30[/tex]
Divide by 6. [tex]|p|\ \textgreater \ -5[/tex]
The solution set is all real numbers because no matter what number is chosen for p, its absolute value will be bigger than any negative number.
6|p|+8 > - 22 ; could be written :
|6.p| + 8 > - 22. Now subtract 8 to both sides
|6p| +8 -8>-22 -8 → |6p| > - 30 . Divide both sides by 6:
|p| > - 5
Clearing the absolute value bars yields:
p> - 5 and p< 5 (remember |x| = a → x = +a and x = -a)
Then :
- 5< p < +5
------------------------(-5)---------------------0--------------------(+5)----------------------
∅============(P)============∅
|6.p| + 8 > - 22. Now subtract 8 to both sides
|6p| +8 -8>-22 -8 → |6p| > - 30 . Divide both sides by 6:
|p| > - 5
Clearing the absolute value bars yields:
p> - 5 and p< 5 (remember |x| = a → x = +a and x = -a)
Then :
- 5< p < +5
------------------------(-5)---------------------0--------------------(+5)----------------------
∅============(P)============∅
